Aptitude Preparation for Campus Placements #3 | Time And Distance | Quantitative Aptitude

Description -
Aptitude Preparation for Campus Placements #3 | Time and Distance | Quantitative Aptitude

This video is made by Akshita Agarwal

Points Covered:
Types of Question
Formula used
Different Questions (each with some new concept)

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Timestamps:
00:00 Types Of Questions
01:00 Question 1
04:07 Question 2
06:04 Question 3
12:41 Question 4
18:07 Outro
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[codestepbystep]

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[dibyojyotideb]

For the 2nd question, the question states that what is the distance that has been covered by each tire. So, the answer should be 8000km itself. 32000km is the distance that is left after the fist tire has covered the journey which is not what the question is asking for.

[devsahani]

Easy solution for question 3:
3x + 4y = 36 and 4x + 3y = 34
Add both eq.
7(x + y) = 70 => x + y = 10
Hence answer is 10

[Ybash]

Honestly your explanation is nice, dono why its so much -ve out here

[nik-njbo]

00:03 Understanding Distance and Time concepts in Aptitude
01:27 Calculating time reduction when speed changes
03:30 Calculating time and distance for empty tires
05:20 Calculating cost and distance for empty tires
07:36 Calculating time and distance using speed
10:00 Understanding time, distance, and equation calculation
14:00 Discussion on time, distance, and equality
16:47 Time and distance calculations differ for time units in seconds and minutes
Crafted by Merlin AI.

[shridharpatil]

mam app sikha rahe ho yad kudko sikha rhe rhe aptitude kyuki eak too itna teda meda likh rhe uprse question hame padhne bhi ni arha or app ho kuch step ese kyu kiya bina bole agge badhe jate hii ....

kare to kre kya bole to bole kya ji

[pruthivirajrao]

mam is cute and just her behaviour while talking is like my sister . Thanks for your time to teach us

[cleweric]

This is great, the quality of questions you are teaching is amazing. I don't know why there is so much negativity in comments. Your point is quite clear how we can use same formula in different conditions with language twist. Please keep up this good work, Thank you!

[mukeshdeore]

Ma'am you're no use in introduction you said that students who don't have any knowledge aptitude will learn from basic and while teaching you're doing calculations on your level. You're not explaining how calculations had been done you directly giving final answers and calculations

[dikhit]

nice videos with proper explanation!
thank you

[Muskanjain-xb]

you explain at your own level…not at our level….it is getting so difficult to understand your point….

[samanthakur]

Not able to mam is not so confident when solving questions 🙂

[dEviL_bIsWaJiT]

Thanks mam for this awesome explanation.
💙

[parth]

Question-2 was really brainstorming
~manan

[MontyBaman-bmck]

why time = t - 25/60 comes in 1st question. We have 30 minutes and 5 minutes in question. There is no explanation about t - 25/60.

[prathmeshplays]

My question 3 and why wrong

Sure! Here's a detailed step-by-step summary of how you approached and solved the problem:

Problem Setup:

Total distance to be traveled: 36 km.

Two parts of the journey (A and B):

Part A: Ram travels at 4 km/h.

Part B: Ram travels at 3 km/h.


In another scenario:

In Part A, Ram travels at 3 km/h.

In Part B, Ram travels at 4 km/h.


The total time taken in both scenarios is equal.


Your Approach:

1. Defining Variables:

Let be the distance traveled in Part A and be the distance traveled in Part B.

In the optimized case:

(distance = speed × time for Part A)

(distance = speed × time for Part B)


In the non-optimized case:

(distance = speed × time for Part A)

(distance = speed × time for Part B)




2. Expressing Times in Terms of Distances:











3. Total Time (2T) in Both Scenarios:

You said the total time is equal to the sum of all individual times for both parts:




2T = t1 + t2 + t3 + t4

This was rewritten as:


2T = \frac{36 - x}{4} + \frac{34 - x}{3} + \frac{x}{3} + \frac{x}{4}

where is the distance covered in Part B.

4. Simplifying the Equation:

You combined the fractions into a single denominator (12):




2T = \frac{3(36 - x) + 4(34 - x)}{12} + \frac{4x + 3x}{12}

This simplifies to:


2T = \frac{3(36) + 4(34) - 3x - 4x + 4x + 3x}{12}

The terms and cancel out.

The equation becomes:


2T = \frac{3(36) + 4(34)}{12}

5. Calculating the Final Expression:

Now, you multiplied the constants:




2T = \frac{108 + 136}{12} = \frac{244}{12}

Simplifying:


2T = 20.33

6. Solving for Time :

To find, you divided by 2:




T = \frac{244}{24} = 1.1667 \, \text{hours}

Conclusion:

The time taken for each scenario is 1.1667 hours, which is the total time for both cases when the total distance traveled is 36 km. This time is the same in both scenarios, as per the problem's condition.

Let me know if you need any further clarifications!

[venkateshkhardekar]

The expression t-25/60 represents the time taken in Case 2, where the student walks at a speed of 6 km/hr.

Here's how it's derived:

1. The student is 25 minutes early when walking at 6 km/hr compared to walking at 5 km/hr.
2. 25 minutes can be converted to hours by dividing by 60, which equals (25/60) hours.
3. So, the time taken in Case 2 is "t" hours (the time taken at 5 km/hr) minus the time gained by being early, which is (25/ 60) hours.

This gives us the expression t-25/60 for the time taken in Case 2.

[naveenkumar-iboe]

In third question why 34*3 becomes 17*1

[zeronpz]

Good Explanation and taking time to create such content is a commendable job!!

[abhijitkumarmanna]

1st question it will be 25mins and t+30 with 5kmph and t+25 for 6kmph, if total time required is 't', the student took 25mins more and 30 mins more time with different speeds as he is late.